我在尝试在 XSLT 中实现 C# 扩展函数时收到以下错误。
Extension function parameters or return values which have CLR type 'Char[]' are not supported.**
代码:
<xsl:variable name="stringList">
<xsl:value-of select="extension:GetList('AAA BBB CCC', ' ')"/>
</xsl:variable>
<msxsl:script language="C#" implements-prefix="extension">
<![CDATA[
public string[] GetList(string str, char[] delimiter)
{
...
...
return str.Split(delimiter, StringSplitOptions.None);
}
]]>
</msxsl:script>
谁能解释这个错误消息以及如何克服它?
编辑:我需要一个解决方案,它仍然可以让我实现拆分功能并利用返回的数组。
谢谢!
最佳答案
XSLT 扩展方法必须返回 XSL 转换中支持的类型。下表显示了 W3C XPath 类型及其对应的 .NET 类型:
W3C XPath Type | Equivalent .NET Class (Type) ------------------------------------------------------ String | System.String Boolean | System.Boolean Number | System.Double Result Tree Fragment | System.Xml.XPath.XPathNavigator Node Set | System.Xml.XPath.XPathNodeIterator
The table is taken from the section Mapping Types between XSLT and .NET in this MSDN Magazine article.
Instead of returning a string[]
array you would have to return an XPathNodeIterator
like it is done in the following example:
<msxsl:script implements-prefix="extension" language="C#">
<![CDATA[
public XPathNodeIterator GetList(string str, string delimiter)
{
string[] items = str.Split(delimiter.ToCharArray(), StringSplitOptions.None);
XmlDocument doc = new XmlDocument();
doc.AppendChild(doc.CreateElement("root"));
using (XmlWriter writer = doc.DocumentElement.CreateNavigator().AppendChild())
{
foreach (string item in items)
{
writer.WriteElementString("item", item);
}
}
return doc.DocumentElement.CreateNavigator().Select("item");
}
]]>
</msxsl:script>
然后在您的 XSL 转换中,您可以使用 xsl:for-each
迭代返回的节点集中的元素:
<xsl:template match="/">
<root>
<xsl:for-each select="extension:GetList('one,two,three', ',')">
<value>
<xsl:value-of select="."/>
</value>
</xsl:for-each>
</root>
</xsl:template>
关于c# - 带有 C# 扩展函数的 XSLT 中的错误消息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3046898/